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Math Help - Series solution for y"+x(e^-x) y = 0

  1. #1
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    Series solution for y"+x(e^-x) y = 0

    Anyone can help me ?

    Question: Find series solution about the origin for
    y"+x(e^-x) y = 0
    and determine the lower bound for the radius of convergence.

    TQ.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by younhock View Post
    Anyone can help me ?

    Question: Find series solution about the origin for
    y"+x(e^-x) y = 0
    and determine the lower bound for the radius of convergence.

    TQ.
    Presumably you know that e^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n so that xe^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}.

    Let y= \sum_{n=0}^\infty a_nx^n so that y"= \sum_{n=0}^\infty n(n-1)a_nx^{n-2} and put that into the differential equation.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Presumably you know that e^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n so that xe^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}.

    Let y= \sum_{n=0}^\infty a_nx^n so that y"= \sum_{n=0}^\infty n(n-1)a_nx^{n-2} and put that into the differential equation.

    But i do not know how to simplify the [ \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1} ] [ \sum_{n=0}^\infty a_nx^n ] .

    thanks for guiding...
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